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3 years ago

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How does the electron structure of atoms change when they form chemical bonds

1 answer:

andriy [413]3 years ago

3 0

For ionic bond
The metal atom will lose electrons to form cations and the non metal atom will gain electron.egNaCl

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If an acid has a ka=1.6x10^-10, what is the acidity of the solution?

LUCKY_DIMON [66]

The information given in the question is not enough to determine the acidity of the solution. This is because, acidity can only be found with the equation: pH = -log [H+].
In order to determine the acidity of the solution, the half titration point value is needed, this will make it possible to determine the value of H30+. If the half point titration value is known, then Ka will be equivalent to pH and the value will be evaluated using the equation: - log (1.6 * 10^-10).

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3 years ago

What are the properties of salt

GrogVix [38]

The compound is sodium chloride

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3 years ago

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In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

SVETLANKA909090 [29]

<u>Answer:</u> The final concentration of potassium nitrate is $5.70\times 10^{-6}M$

<u>Explanation:</u>

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$ .......(1)

<u>Calculating for first dilution:</u>

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL$

Putting values in equation 1, we get:

$7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M$

<u>Calculating for second dilution:</u>

$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

$1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M$

Hence, the final concentration of potassium nitrate is $5.70\times 10^{-6}M$

8 0

3 years ago

True or False: As the ice warms up, the molecules get more energy and move around more from place to place to form liquid water.

Arada [10]

The answer is TRUE.

Hope it helps.

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3 years ago

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In a metabolic pathway, succinate dehydrogenase catalyzes the conversion of succinate to fumarate. the reaction is inhibited by

Pani-rosa [81]

Competitive inhibitor

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3 years ago